# If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM

**Solution:**

We know if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

This is referred as SAS criterion for two triangles.

Given, ΔABC ∼ ΔPQR

⇒ ∠ABC = ∠PQR (corresponding angles) --------- (1)

⇒ AB/PQ = BC/QR (corresponding sides)

⇒ AB/PQ = (BC/2) / (QR/2)

⇒ AB/PQ = BD/QM (D and M are mid-points of BC and QR) ------------ (2)

In ΔABD and ΔPQM,

∠ABD = ∠PQM (from 1)

AB/PQ = BD/QM (from 2)

⇒ ΔABD ∼ ΔPQM (SAS criterion)

⇒ AB/PQ = BD/QM = AD/PM (corresponding sides)

⇒ AB/PQ = AD/PM

Hence proved.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 Question 16

**Summary:**

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, it is proved that AB/PQ = AD/PM.

**☛ Related Questions:**

- In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ΔABC ~ ΔAMP (ii) CA/PA = BC/MP.
- CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that: (i) CD/GH =AC/FG (ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF
- In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥AC, prove that ∆ ABD ~ ∆ ECF.
- Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR.

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